Fewpal
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An open belt running over two pulleys 24cm and 60cm diameters. Connects two parallel shaft 3m apart and transmits 3.75KW from the smaller pulley that rotates at 300RPM, µ = 0.3, and the safe working tension in 100N/cm width. Determine 

(i) Minimum width of the belt. 

(ii) Initial belt tension. 

(iii) Length of the belt required.

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Given that, 

D1 = 60cm 

D2 = 24cm 

N2 = 300rpm µ = 0.3 

X = 3m = 300cm 

P = 3.75KW

Safe Tension = Maximum tension = 100N/cm width = 100b N b = width of belt 

Tmax = 100b  ...(i)

Let θ = Angle of contact

Sinα = (r1 – r2)/X = (30 – 12)/300 ; α = 3.45°, ...(ii)

 = Π – 2α = (180 – 2 X 3.45) = 173.1°

= (173.1°)X Π/180 = 3.02 rad ......(iii)

Now,

Using the relation, Ratio of belt tension = T1/T2 = eµθ = e(0.3)(3.02)

T1 = 2.474 T2 ......(iv)

Now,

V = πDN/60 m/sec, D is in meter and N is in RPM

 = 3.14 X (0.24)(300)/60 = 3.77m/sec ......(v)

Power Transmitted (P) = (T1 – T2).v/1000 Kw

3.75 = (T1 – T2)X 3.77/1000

T1 – T2 = 994.7N .......(vi)

 From relation (iv) and (v), we get:

T1 = 1669.5N ...........(vii)

T2 = 674.8N ...........(viii)

(i) For width of the belt 

But T1 = Tmax = 100b; 1669.5 = 100b; b = 16.7cm 

(ii) For initial tension in the belt 

Let To = initial tension in the belt

To = (T1 + T2)/2

= (1669.5 + 674.8)/2

To = 1172.15N

(iii) For length of belt

Putting all the value, we get

L = 7.33m.

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